A342304 k-digit positive numbers exactly one of whose substrings is divisible by k.

1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 23, 25, 27, 29, 41, 43, 45, 47, 49, 61, 63, 65, 67, 69, 81, 83, 85, 87, 89, 101, 104, 107, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122, 125, 128, 131, 134, 137, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 152, 155, 158, 161

Offset: 1

Author

Bernard Schott, Mar 08 2021

Keywords

• nonn
• base

Comments

Inspired by the 413th problem of Project Euler (see link) where such a number is called "one-child number".

There are k*(k+1)/2 substrings. All are considered, even when some are duplicates as strings or as numbers (see the Example section). 0 is always divisible by k so any number with two or more 0 digits is not a term. - Kevin Ryde, Mar 08 2021

The 2-digit terms are odd.

The number of k-digit terms for k = 1, 2, 3 is respectively 9, 20, 360.

From Robert Israel, Mar 11 2021: (Start)

5-digit terms are numbers starting with 5, and with no other digits 5 or 0.

There are no 10-digit terms. (End)

Examples

			107 is a 3-digit one-child number since among its substrings 1, 0, 7, 10, 07, 107 only 0 is divisible by 3.
222 is a 3-digit one-child number since among its substrings 2, 2, 2, 22, 22, 222 only 222 is divisible by 3.
572 is not a 3-digit one-child number, since among its substrings 5, 7, 1, 57, 72, 572 both 57 and 72 are divisible by 3.
616 is not a 3-digit one-child number, since among its substrings 6, 1, 6, 61, 16, 616 the two 6's are both divisible by 3.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Project Euler, Problem 413: One-child Numbers.

Crossrefs

Cf. A063527.

Programs

Maple

filter:= proc(n) local L,d,i,j,k, ct, x;
  L:= convert(n,base,10);
  d:= nops(L);
  ct:= 0:
  for i from 1 to d do
    for j from i to d do
      x:= add(L[k]*10^(k-i),k=i..j);
      if x mod d = 0 then ct:= ct+1; if ct = 2 then return false fi fi;
  od od;
  evalb(ct = 1)
end proc:
select(filter, [$1..200]); # Robert Israel, Mar 11 2021

Python

def ok(n):
    s, c = str(n), 0
    ss = (s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1))
    for w in ss:
        if int(w)%len(s) == 0: c += 1
        if c == 2: return False
    return n > 0 and c == 1
print([k for k in range(162) if ok(k)]) # Michael S. Branicky, Aug 15 2022